The indefinite integral can be computed and there arises an \(F_1\)  Appell function [1]}.
\(I=\dfrac{\Gamma (n+\alpha +1)}{\Gamma (n+1)\Gamma (\alpha +1)}\sum _{i=0}^n\dfrac{(-n)_i(n+\alpha +\beta +1)_i}{(\alpha +1)_i}\dfrac{1}{i!}\left(\dfrac{1}{2}\right)^i \\\left[-\dfrac{2^\beta }{(\alpha +1+i)}\dfrac{(1-t)^{\alpha +1+i}}{(1-x)^\lambda }F_1\left(\begin{array}{l}\alpha +1+i,-\beta ,\lambda \\\alpha +2+i\end{array};\dfrac{1-t}{2},\dfrac{1-t}{1-x}\right)\right]^1_x\) 
